A 150.0 kg rocket moving radially outward from Earth has a...

A 150.0 kg rocket moving radially outward from Earth has a speed of 3.70 $\mathrm{km} / \mathrm{s}$ when its engine shuts off 200 $\mathrm{km}$ above Earth's surface. (a) Assuming negligible air drag acts on the rocket, find the rocket's kinetic energy when the rocket is 1000 $\mathrm{km}$ above Earth's surface. (b) What maximum height above the surface is reached by the rocket?

A 150.0 kg rocket moving radially outward from Earth has a speed of 3.70 $\mathrm{km} / \mathrm{s}$ when its engine shuts off 200 $\mathrm{km}$ above Earth's
surface. (a) Assuming negligible air drag acts on the rocket, find
the rocket's kinetic energy when the rocket is 1000 $\mathrm{km}$ above
Earth's surface. (b) What maximum height above the surface is
reached by the rocket?

Key Concepts

Conservation of Mechanical Energy

This principle states that in the absence of non-conservative forces (like friction or air drag), the total mechanical energy (the sum of kinetic and potential energy) of a system remains constant. It allows one to equate the energy at one position to the energy at another, making it possible to solve for unknown quantities such as the speed or height of an object moving in a gravitational field.

Gravitational Potential Energy in a Radial Field

In a radially symmetric gravitational field, the potential energy is given by the formula U = -GMm/r, where G is the gravitational constant, M is the mass of the Earth, m is the mass of the object, and r is the distance from the center of the Earth. This concept is crucial for analyzing motions like the one in the problem, where changes in altitude lead to changes in potential energy.

Kinetic Energy

Kinetic energy is defined as the energy an object possesses due to its motion and is calculated using the formula K = ½mv². Understanding kinetic energy is essential when using energy conservation to determine the behavior of objects moving under the influence of gravitational fields, particularly when computing changes in speed as the object moves to different altitudes.

Transcript

00:01 So here we can say that we know that the radius of the earth is equaling 6 .37 times 10 to the 6th meters.

00:09 We know that the mass of the earth is equaling 5 .98 times 10 to the 24th kilograms.

00:18 And we also know that here the orbital radius of the rocket would be equal to the radius of the earth plus the altitude of the rocket.

00:25 This is equaling 6 .57 times 10 to the 6 meters.

00:33 We know that our initial rather would be where the our initial would be a 200 kilometers above the earth surface where the engine shuts off and then we're trying to find the rocket's kinetic energy 1 ,000 kilometers above the earth surface.

01:05 So this is where r not would be the point where the engine shuts off.

01:10 And then r would be equal to 6 .37 times 10 to the 6 meters plus 1 ,000 kilometers.

01:19 So 7 .37 times 10 to the 6 meters.

01:24 Now we have that due to the law of conservation of energy for part a, we can say that the initial kinetic energy plus the initial potential equals the final kinetic energy plus the final potential, we can then say that this is one half m times 3 .7 0 times 10 to the third meters per second quantity squared minus this would be g times m times m divided by r not.

01:58 This would be equal to k minus g times m times m.

02:06 Over r.

02:08 Now, we can plug in for the mass of the rocket, the mass of the earth, the radius, the, rather the height at which the engine shuts off, and the height of 1 ,000 kilometers above the earth's surface.

02:23 You can use your t -i -84, 85, or 89 in order to solve for k...

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